How to prove this relationship for Pearson´s c.c.


where $(X_{1},Y_{1}),(X_{2},Y_{2})$ are i.i.d. as $(X,Y)$ and $E(X)=\mu, E(Y)=\nu.$

Thanks in advance for any comments.


Try adding and subtracting the means and then expand the squares and the products, maybe work each expectation apart to see the result. For instance, $$E(X_1-X_2)^2=E\big((X_1-\mu)-(X_2-\mu)\big)^2=$$ $$=E(X_1-\mu)^2-2E(X_1-\mu)(X_2-\mu)+E(X_2-\mu)^2=\cdots$$ and so on. Try to check that in this case you get $2 V(X)$.

  • 1
    $\begingroup$ Thank you so much, it rally works. Just need to expand denominator/numerator and use independence of components of these vectors. $\endgroup$ – stanly Nov 30 '17 at 13:25

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